3
$\begingroup$

I am sorry if this is a basic question, but I don't think in MSE I will receive any answers.

Let $(M^3,g)$ be a compact and oriented Riemannian $3$-manifold. Let $\alpha$ and $\beta$ be the integral currents supported on the compact, connected, oriented embedded surfaces $\Sigma_1$ and $\Sigma_2$ with multiplicities $\theta_1$ and $\theta_2$, that is:

$$ \alpha(\omega) = \int_{\Sigma_1} \langle \omega(x), \tau_1 (x) \rangle \theta_1(x) d \mathcal{H}^2(x), \quad \omega \in \Omega_2(M), $$

and similarly for $\beta$. Here, $\tau_1$ is a $2$-plane field that orients $\Sigma_1$ and $\theta_1$ is an integrable positive integer-valued function on $\Sigma_1$.

Assume that $\Sigma_1$ and $\Sigma_2$ are disjoint. Is it true that the mass of $\alpha$ is smaller than the mass of the sum $\alpha + \beta$?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

Yes, this is true, from the fact that the supports of these currents are disjoint. In fact, I think the same argument should work for two general currents, as long as their supports have disjoint neighborhoods.

Write $M(\alpha)$ and $M(\alpha + \beta)$ for the masses of $\alpha$ and $\alpha + \beta$, respectively.

Then there is a sequence of 2-forms $\omega_k$ such that $|\omega_k|_{C^0(M)} = 1$ for every $k$ and $$|\alpha(\omega_k)| \to M(\alpha).$$

Let $\chi: M \to \mathbb{R}$ be a smooth, non-negative cutoff function that is equal to $1$ in a neighborhood of $\Sigma_1$ and $0$ in a neighborhood of $\Sigma_2$. Then it follows that $$\alpha(\omega_k) = \alpha(\chi\omega_k)$$ for every $k$, and moreover $$\beta(\chi\omega_k) = 0$$ for every $k$.

It follows that $$|(\alpha + \beta)(\chi\omega_k)| \to M(\alpha).$$

However, by definition, $$|(\alpha + \beta)(\chi\omega_k)| \leq M(\alpha + \beta)$$ for every $k$.

It follows from the above two statements that we must have the inequality $$M(\alpha) \leq M(\alpha + \beta).$$

Improve this answer
$\endgroup$
2
  • $\begingroup$ Nice argument. In the particular case I described, the mass of $\alpha$ is stricly smaller than that of the sum, right? $\endgroup$
    – Eduardo Longa
    Commented Aug 13, 2020 at 17:47
  • $\begingroup$ Oh, I wasn't very careful about that. I think my argument can be fixed to show that, by picking some two-form $\eta$ supported disjointly from $\chi\omega_k$ such that $\beta(\eta) > 0$. Then, if we assume $\alpha(\chi\omega_k) > 0$ for every $k$ as well, you get $\lim_{k \to \infty} |(\alpha + \beta)(\chi\omega_k + \eta)|$ is equal to $\lim_{k \to \infty} |\alpha(\chi\omega_k)| + |\beta(\eta)|$. This is strictly greater than $M(\alpha)$, but bounded above by $M(\alpha + \beta)$ which gives you the strict inequality. $\endgroup$
    – Rohil Prasad
    Commented Aug 13, 2020 at 17:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .